Mathhelp346
log base 6 [log base 5 (log base 3 x)] = 0



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Mathhelp346
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\[\log_{6} [\log_{5} (\log_{3}x) ]=0 \]

anonymous
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pealing off one at a time, this tells you
\[\log_5(\log_3(x))=1\] as a first step

anonymous
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this in turn means \(\log_3(x)=5\)

anonymous
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and finally this gives \(x=3^5\)

Mathhelp346
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Wait, how did you know that
\[\log_{5} (\log_{2} (x))=1\] ?

Mathhelp346
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And what happened to\[\log_{6}\]

Mertsj
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\[\log_{a}x=b \]
means
\[a ^{b}=x\]

Mertsj
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The great one used that principle 3 times.

Mathhelp346
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The great one...haha. :P